Omid - Challenge 344

data-challenges
advanced-exercises
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Published

March 24, 2026

Illustration for Omid - Challenge 344

Challenge Description

πŸ”° 🧩 Adding Explanations If you want to include explanations or extra notes: 🌐 Sharing External Content πŸ—£ Feedback I always appreciate your feedback β€” feel free to share an…

Solutions

library(tidyverse)
library(readxl)

path <- "300-399/344/CH-344  Matrix Calculation.xlsx"
input1 <- read_excel(path, range = "B4:F8", col_names = FALSE) %>% as.matrix()
input2 <- read_excel(path, range = "B14:F18", col_names = FALSE) %>% as.matrix()
test1 <- read_excel(path, range = "H4", col_names = FALSE) %>% pull()
test2 <- read_excel(path, range = "H14", col_names = FALSE) %>% pull()

is_symmetric <- function(mat) {
  all(mat == t(mat))
}
max_sym1 <- max(which(map_lgl(1:min(dim(input1)), function(x) {
  is_symmetric(input1[1:x, 1:x])
})))
max_sym2 <- max(which(map_lgl(1:min(dim(input2)), function(x) {
  is_symmetric(input2[1:x, 1:x])
})))

all.equal(test1, max_sym1) # True
all.equal(test2, max_sym2) # True

  • Logic:

    • Reads the workbook ranges needed for the challenge

  • Strengths:

    • The R solution stays close to the workbook rule and keeps the transformation compact.

  • Areas for Improvement:

    • The code assumes the sheet structure and source ranges remain stable.

  • Gem:

    • The strongest part of the solution is choosing the right intermediate representation before shaping the final output.
import numpy as np
import pandas as pd

path = "300-399/344/CH-344  Matrix Calculation.xlsx"

input1 = pd.read_excel(path, usecols="B:F", skiprows=3, nrows=5, header=None).values
input2 = pd.read_excel(path, usecols="B:F", skiprows=13, nrows=5, header=None).values
test1 = pd.read_excel(path, usecols="H", skiprows=3, nrows=1, header=None).iloc[0,0]
test2 = pd.read_excel(path, usecols="H", skiprows=13, nrows=1, header=None).iloc[0,0]

def is_symmetric(mat):
    return np.all(mat == mat.T)

max_sym1 = max([x for x in range(1, min(input1.shape)+1) if is_symmetric(input1[:x, :x])])
max_sym2 = max([x for x in range(1, min(input2.shape)+1) if is_symmetric(input2[:x, :x])])

print(np.isclose(test1, max_sym1))  # True
print(np.isclose(test2, max_sym2))  # True

  • Logic:

    • Reads the workbook ranges needed for the challenge

    • Applies the rule iteratively until the output stabilizes

  • Strengths:

    • The Python version follows the same rule in a direct dataframe-oriented implementation.

  • Areas for Improvement:

    • The code assumes the workbook layout remains stable, so any sheet redesign would require small adjustments.

  • Gem:

    • The implementation stays close to the original workbook rule instead of adding unnecessary abstraction.

Difficulty Level

This task is moderate:

  • The business rule is readable, but the workbook still requires careful implementation to reach the expected layout.